Finding the max frequency of a driven oscillator

  • #1
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Main Question or Discussion Point

So I've derived the equation for the amplitude of a driven oscillator as:

[itex]\huge A=\frac{F}{m\sqrt{(\omega_{0}^{2}-\omega_{d}^{2})^{2}+4\gamma^{2}\omega_{d}^{2}}}[/itex]

Which is what my lecturer has written. Then taking the derivative and setting it to 0 to get the turning point. He makes this leap:

https://imgur.com/a/gE7Y0Di



How does he do that? I can't do it.

Also an auxiliary question. I was watching Walter Lewin on this here .
And he uses

[itex]\huge
{\gamma}=\frac{b}{m}[/itex]

Whereas I've been taught:

[itex]\huge
{\gamma}=\frac{b}{2m}[/itex]

Where ϒ is the damping coefficient. Which is correct?
 

Answers and Replies

  • #2
Charles Link
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His definition is ok. Both are used for ## \gamma ##. ## \\ ## If you take the derivative ## \frac{dA}{d \omega_D} ## and set it equal to zero, you get the peak of the ## A ## vs. ## \omega_D ## graph,(the resonant frequency), which peaks just slightly off from ## \omega_o=\sqrt{k/m} ##. The derivative is quite a simple operation with the chain rule. ## \\ ## Note: He starts with ## F(t)=Fe^{i \omega_D t} ##, and computes the amplitude ## A=|x| ##.
 

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